Hwksolns10TA - Solutions to HW 10 Problems and Exercises:...

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Solutions to HW 10 Problems and Exercises: 37.28. Visualize: At t t t 0 s, the origins of the S, S , and S reference frames coincide. Solve: We have 11 22 1 ( / ) 1 (0.80) 1.667. vc     (a) Using the Lorentz transformations, 8 ( ) 1.667 1200 m (0.80)(3 10 m/s)(2.0 10 s) xx v t 6 1200 m  8 6 2 2 8 (0.80)(3 10 m/s)(1200 m) 1.667 2.0 10 s 2.0 s 31 0 m / s vx tt c      (b) Using v  c , the above equations yield  x 2800 m and 8.67 s. t   37.29. Solve: We have 1 1 0.60 1.25. In the earth’s reference frame, the We assume that the axes of the two frames coincide at t=t’= 0 sec Lorentz transformations yield   10 8 10 10 81 0 2 2 8 ( ) 1.25 3.0 10 m 0.60 3.0 10 m/s 200 8.25 10 m 8.3 10 m 0.60 3.0 10 m/s 3.0 10 m 1.25 200 325 s 330 s 3.0 10 m/s v t vx c     37.30. Model: S is the ground’s frame and S is the rocket’s frame. S moves with velocity v 0.5 c relative to S. Solve: (a) We have 1 / 1 0.50 1.155. Applying the Lorentz transformations to the lightning strike at x 0 m and t 10 s,       85 5 2 (1.155) 0 m 0.5 3.0 10 m/s 1 10 s 1732 m 1700 m (1.155)(1 10 s 0 s) 11.55 s 12 s v t vx c     For the lightning strike at x 30 km and t 10 s,     48 5 84 5 2 8 1.155 3.0 10 m 0.50 3.0 10 m/s 1 10 s 32.91 m 33 m 0.50 3.0 10 m/s 3.0 10 m 1.155 1 10 s 46.2 s 46 s x t (b) The events in the rocket’s frame are not simultaneous. The lightning is observed to strike the pole before the tree by 46 12 58 s. Hence we could infer, from direct application of Lorentz transformation (which is less confusing and more universal ), that simultaneity is relative.
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37.31. Model: The rocket and the earth are inertial frames. Let the earth be frame S and the rocket be frame S . S moves with v 0.8 c relative to S. The bullet’s velocity in reference frame S is u 0.9 c . Solve: Using the Lorentz velocity transformation equation,  22 0.9 0.8 0.36 1/1 0 . 9 0 . 8 / uv c c uc uv c c c c      The bullet’s speed is 0.36 c along the x -direction. Note that the velocity transformations use velocity , which can be negative, and not speed. Hence it is absolutely important to define the positive direction of motion. Is most cases the direction of S’ w.r.t S is taken to be positive for bothe reference frames S and S’ 37.32. Model: The proton and the earth (the Laboratory) are inertial frames. Let the earth be frame S and the proton be frame S . S moves with v 0.9 c . The electron’s velocity in the laboratory frame is 0.9 c . Notice that the direction of motion of proton frame S’ w.r.t lab frame S is towards right. This direction is taken to be positive for judging velocities in both S and S’ frames.
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Hwksolns10TA - Solutions to HW 10 Problems and Exercises:...

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